Calculate A Sgurr for a reaction for which AH - 1575 kJ at 25°C. Assume constant...

Question

Question

Answer 1

Answer #1

Answer:

$\Delta$ S_surr = 5.29 × 10³ J/K

Explanation:

At constant temperature and pressure, the formula for $\Delta$ S_surr is,

Assura = a Hsuror here, AS suror as of surroundings Aitsuror = AH of surroundings. T = Temperature

In the question it is given that,

T = 25°C = 298.15 K

$\Delta$ H_rxn = -1575 kJ

$\Delta$ H_rxn is the $\Delta$ H of system or $\Delta$ H_sys.

So, $\Delta$ H_sys = -1575 kJ

$\Delta$ H_sys + $\Delta$ H_surr = 0

$\Delta$ H_surr = - $\Delta$ H_sys

So, $\Delta$ H_surr = -(-1575 kJ) = 1575 kJ

$\Delta$ H_surr = 1575 kJ

T = 298.15 K

Now, put these values in the formula for $\Delta$ S_surr,

Assura = 15 75 kJ 298.15 K os suror = 5,29 kJ/ (:1 kJ = 1+ 103J) Tos saree = 5.29 * 103 J/K !

Answer 2

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